We have been given that :-
The length of a parasite in experiment A is [tex]2.5 \times 10^{-3}[/tex]
The length of a parasite in experiment B is [tex]1.25 \times 10^{-4}[/tex]
Let us write the the length of the parasite in experiment A in the exponent of -3.
[tex]1.25 \times 10^{-4}= 0.125 \times 10^{-3} [/tex]
Clearly, the length of parasite in experiment A is greater than the length of parasite in experiment B.
The difference in the length is given by
[tex]2.5 \times 10^{-3} -0.125 \times 10^{-4}[/tex]
[tex]=2.375 \times 10^{-3}[/tex]
Therefore, the length of the parasite in experiment A is [tex]=2.375 \times 10^{-3}[/tex] inches greater than the length of the parasite in experiment B.
Answer: 20 times
Step-by-step explanation:
Given : The length of a parasite in experiment A = [tex]2.5\times10^{-3}\text{ inch}[/tex]
The length of a parasite in experiment B = [tex]1.25\times10^{-4}\text{ inches}[/tex]
Then, the number of times the length of the parasite in experiment A is greater as compared to the length of the parasite in experiment B:-
[tex]n=\dfrac{2.5\times10^{-3}}{1.25\times10^{-4}}\\\\=\dfrac{2.5}{1.25}\times10^{-3-(-4)}\\\\=2\times10^{-3+4}=2\times10^{1}=20[/tex]
Hence, the length of the parasite in experiment A is 20 times greater as compared to the length of the parasite in experiment B
